Question

A random sample of n-15 heat pumps of a certain type yielded the following observations on lifetime (in years): 2.0 1.5 6.0 1.9 5.4 0.4 1.0 5.3 15.6 0.9 4.8 0.9 12.3 5.3 0.6 (a) Assume that the lifetime distribution is exponential and use an argument parallel to that of this example to obtain a 95% CI for expected (true average) lifetime. (Round your answers to two decimal places.) ) years (b) How should the interval of part (a) be altered to achieve a confidence level of 99%? 0 A 99% confidence level requires using critical values that capture an area of 0.1 in each tail of the chi-squared distribution. 0 A 99% confidence level requires using a new value of n to capture an area of 0.005 in each tail of the chi-squared distribution. 0 A 99% confidence level requires using critical values that capture an area of 0.005 in each tail of the chi-squared distribution. 0 A 99% confidence level requires using a new value of n to capture an area of 0.1 in each tail of the chi-squared distribution. (c) what is a 95% CI for the standard deviation of the lifetime distribution? [Hint: What is the standard deviation of an exponential random variable?] (Round your answers to two decimal places.) years

Answer 1

a)

Calculate the 95% confidence interval for the expected lifetime. Assume the lifetime distribution is an exponential distribution with mean x, ;λ)-2ΑΣΧ,has a χ2 distribution with 2n degrees of freedom. h(x Consider Then the 95% Confidence interval for is given by, 2n 22x From the table values, (975.0) 16.791 and (023s.0) 46.979 (0025.30)

xi-63.9

(16.791/127.8 < lambda < 46.979/126.4)
hence 95% confidence interval is
(127.8/46.979 < 1/lambda < 127.8/16.791)
= ( 2.72 ,7.61 )

b)

option C) is correct

c)

answer will be same as a) part as mean = sd in exponential distribution

= ( 2.72 ,7.61 )

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